1. SPRING-MASS OSCILLATORS AP Physics Unit 8

2. Recall Hooke’s Law Applied force (F applied ) stretches or compresses spring from its natural length Restoring force (F r ) acts to return spring to lowest energy state

3. An energy approach to SHM Stretched/compressed spring stores elastic potential energy: U s = ½kx 2 When released, mass oscillates about its equilibrium position as PE  KE  etc –Amplitude of oscillation is x max –At x=0, U s = 0 so K is maximized –At x=A, U s is maximized, so K=0 U s, max = ½kA 2

4. Example #1 A 2.0 kg block is attached to an ideal spring with a force constant of 500 N/m. The spring is stretched 8.0 cm and released. When the block is 4.0 cm from equilibrium –what is the total energy of the system? –what is the velocity of the block? By energy conservation, E=1.6J at every spring position

5. An energy approach to SHM Since energy is conserved and –at x=A, U s is maximized and K=0 –at x=0, U s = 0 so K is maximized it follows that therefore

6. SHM and the Reference Circle Motion of the shadow cast by a particle moving in a vertical circle mimics SHM –Amplitude corresponds to the radius of the circle –Period of the oscillation corresponds to the period of the UCM or since

7. Example #2 A 2.0kg block is attached to a spring with a force constant of 300 N/m. Determine the period and frequency of the oscillations.

8. Vertical Spring-Mass Oscillators As it turns out, the behavior is the same regardless of the orientation, i.e. gravity does not affect the period or frequency of the oscillations. Sounds improbable? Let’s see why it is not…

9. Vertical SMOs Consider a spring with constant k on which a mass m is hung, stretching the spring some distance x The spring is in equilibrium: F applied = F r or kx=mg If the spring is further displaced by some amount A, the restoring force increases to k(x+A) while the weight remains mg

10. Vertical SMOs The net force on the block is now F= k(x+A)- mg But since kx=mg, the force on the block is F= kA. This is Hooke’s Law! Instead of oscillating about the natural length of the spring as happens with a horizontal SMO, oscillations of a vertical SMO are about the point at which the hanging mass is in equilibrium!

11. Example #3 A 1.5 kg block is attached to the end of a vertical spring with a constant of 300 N/m. After the block comes to rest, it is stretched an additional 2.0 cm and released. –What is the frequency of the oscillation? –What are the maximum & minimum amounts of stretch in the spring? Since A=2.0 cm, the spring is stretched a maximum of 6.9 cm and a minimum of 2.9 cm